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AP Statistics 4.8 Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Means- Exam Style Questions - MCQs - New Syllabus

Question 

A 95% confidence interval for \(\mu_A-\mu_B\) is \((-4.5,\;5.2)\). Based on this information, which of the following is an appropriate conclusion? Assume all conditions for inference are met.

(A) We have sufficient evidence to suggest that \(\mu_A > \mu_B\)
(B) We have sufficient evidence to suggest that \(\mu_A < \mu_B\)
(C) We have sufficient evidence to suggest that \(\overline{x}_A < \overline{x}_B\)
(D) We do not have sufficient evidence to suggest that \(\mu_A\) is different from \(\mu_B\)
(E) We do not have sufficient evidence to suggest that \(\overline{x}_A\) is different from \(\overline{x}_B\)

▶️ Answer/Explanation

The 95% confidence interval for the difference in population means is:

\(\mu_A-\mu_B=(-4.5,\;5.2)\)

Notice that the interval contains 0. This means that a difference of zero between the population means is a plausible value based on the sample data.

Because 0 lies within the confidence interval, we do not have convincing evidence that the population means are different. Therefore, we fail to conclude that \(\mu_A\) and \(\mu_B\) differ.

The conclusion must be about the population means, not the sample means, so options involving \(\overline{x}_A\) and \(\overline{x}_B\) are inappropriate.

Answer: (D)

Question 

In a survey of 500 students, it was found that 200 students only own a smartphone, 150 students only own a tablet, and 100 students own both a smartphone and a tablet. If one of the surveyed students is chosen at random, what is the probability that the student owns a tablet given that the student already owns a smartphone.

(A) \( \frac{100}{200} \)

(B) \( \frac{100}{300} \)

(C) \( \frac{100}{500} \)

(D) \( \frac{150}{300} \)

(E) \( \frac{150}{500} \)

▶️ Answer/Explanation

We are asked to find:
\( P(\text{Tablet} \mid \text{Smartphone}) \)

Using the conditional probability formula:
\( P(A \mid B)=\frac{P(A \cap B)}{P(B)} \)

Number of students who own both a smartphone and a tablet:
\( 100 \)

Number of students who own a smartphone:
\( 200 + 100 = 300 \)

Therefore:
\( P(\text{Tablet} \mid \text{Smartphone}) \)
\( = \frac{100}{300} \)
\( = \frac{1}{3} \)

Thus, the correct answer choice is the fraction \( \frac{100}{300} \).

Answer: (B)

Question 

The marketing director for an ice cream company investigated whether there was a difference in preference for two new ice cream flavors—cotton candy and mango. Each participant from a large group of people was randomly assigned to taste one of the two flavors. After tasting, each person rated the flavor on a numerical scale from 1 to 5, where 1 represented strongly dislike and 5 represented strongly like. A two-sample t-interval for a difference between means (cotton candy minus mango) was constructed. Based on the interval, there was convincing statistical evidence of a difference in population mean flavor ratings, with mango having the greater sample mean rating. Which of the following could be the constructed interval?

(A) \((-20,-15)\)
(B) \((-2.1,-1.3)\)
(C) \((-1.4,2.6)\)
(D) \((1.5,2.7)\)
(E) \((15,20)\)

▶️ Answer/Explanation

The interval is for:

\(\mu_{\text{cotton candy}}-\mu_{\text{mango}}\)

Since mango has the greater sample mean rating, the difference (cotton candy − mango) should be negative. Also, because there is convincing statistical evidence of a difference, the confidence interval must not contain 0.

Choices (A) and (E) are impossible because ratings are on a scale from 1 to 5, so differences cannot be that large. Choice (C) contains 0, so it does not provide convincing evidence of a difference. Choice (D) is entirely positive, indicating cotton candy would have the greater mean rating. Only choice (B) is entirely negative and excludes 0.

Answer: (B)

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